Problem: In the NBA in 2003, Yao Ming was one of the tallest players at $7'5''$. Earl Boykins was the shortest player at $5'5''$. How many inches taller than Boykins was Ming?
Answer: The difference between $7'5''$ and $5'5''$ is exactly two feet. Since there are twelve inches in one foot, two feet equal $2\;\cancel{\text{feet}}\times 12\;\dfrac{\text{inches}}{\cancel{\text{foot}}} = \boxed{24}$ inches.